Characteristic not implies isomorph-free in finite group

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties, when the big group is a finite group. That is, it states that in a finite group, every subgroup satisfying the first subgroup property (i.e., characteristic subgroup) need not satisfy the second subgroup property (i.e., isomorph-free subgroup)
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Statement

Statement with symbols

There exists a finite group $G$ and a characteristic subgroup $H$ of $G$ such that $H$ is not an isomorph-free subgroup of $G$. In other words, there exists another subgroup $K$ of $G$ that is isomorphic to $H$.

Related facts

• Characteristic not implies isomorph-containing
• Characteristic not implies sub-isomorph-free in finite
• Characteristic not implies isomorph-normal in finite
• Characteristic not implies sub-(isomorph-normal characteristic) in finite
• Characteristic not implies injective endomorphism-invariant

Proof

Example of the dihedral group

Further information: dihedral group:D8, subgroup structure of dihedral group:D8, center of dihedral group:D8

Let $G$ be the dihedral group of order eight, given as follows, where $e$ denotes the identity element of $G$:

$G = \langle a,x \mid a^4 = x^2 = e, xax = a^{-1} \rangle$.

Let $H$ be the center of $G$. $H$ is a subgroup of order two generated by $a^2$.

• $H$ is characteristic.
• $H$ is not isomorph-free: The subgroup $\langle x \rangle$ of $G$ is isomorphic to $H$.